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解方程x/x+1 +1 = 2x+1/x

解:两边同时乘以x(x+1) 的x^2+x(x+1)=2x^2(x+1)+(x+1) 化简得2x^3=-1 x^3=-1/2 x=(-1/2)^(1/3)

换元法 解: 设(x+1)/x²=t,则x²/(x+1)=1/t,于是 t-2/t=1 t²-2=t t²-t-2=0 (t+1)(t-2)=0 t=-1或2 于是 (x+1)/x²=-1或(x+1)/x²=2 -x²=x+1或2x²=x+1 x²+x+1=0或2x²-x-1=0 x²+x+1=0无解 2...

解:1.x-1>0,即x>1时 x-1=-2x+1 解得x=2/3,与x>1矛盾,舍去 2.x

1、(2x-1)/(x-3)=(2x-1)/(x+1), ——》(2x-1)/(x-3)-(2x-1)/(x+1)=4(2x-1)/[(x-3)(x+1)]=0, ——》2x-1=0, ——》x=1/2; 2、1/(2-x)=1/(x-2)-(6-x)/(3x^2-12), ——》1/(x-2)-(6-x)/(3x^2-12)-1/(2-x)=(7x+6)/(3x^2-12)=0, ——》7x+6=0, ——》x=-6/7。

解: 1/(x+1)=2/(x-1) 方程两边同时乘(x+1)(x-1) x-1=2(x+1) x-1=2x+2 x-2x=2+1 -x=-3 x=-3 经检验x=-3是原方程的根 2x/(2x+5)+5/(5x-2)=1 方程两边同时乘(2x+5)(5x-2) 2x(5x-2)+5(2x+5)=(2x+5)(5x-2) 10x²-4x+10x+25=10x²+21x-10 -4x...

1/(x+2)+1/(2x-1)=0 通分 【(2x-1)+(x+2)】/【(x+2)(2x-1)】=0 3x+1=0 x=-1/3

令t=(x+1)/x^2 方程化为;t-2/t=1 即t^2-t-2=0 (t-2)(t+1)=0 t=2, -1 当t=2时,2x^2-x-1=0, 得:(2x+1)(x-1)=0 得:x=-1/2, 1 当t=-1时,x^2-x-1=0, 得:x=(1+√5)/2, (1-√5)/2 经检验,共有以上4个根

(2x+1)/x-3x/(2x+1)=2 (2x+1)^2-3x^2=2x(2x+1) (2x+1)^2-2x(2x+1)-3x^2=0 (2x+1-3x)(2x+1+x)=0 (1-x)(3x+1)=0 x1=1 x2=-1/3

(x-1)/x-2x/(x-1)=-1 两边乘以x(x-1)得(x-1)²-2x²=-x(x-1) 整理得3x=1 解得x=1 或移项 (x-1)/x=2x/(x-1)-1=(x+1)/(x-1) 即得(x-1)²=x(x+1) 整理得3x=1 解得x=1

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